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CIE 9709 2023 June Paper 61 Q5

A Level / CIE / S2

CIE 9709 2023 June Paper 61 Paper · Question 5

题目

Problem

Large packets of rice are packed in cartons, each containing 2020 randomly chosen packets. The masses of these packets are normally distributed with mean 1010 g1010\,\mathrm{g} and standard deviation 3.4 g3.4\,\mathrm{g}. The masses of the cartons, when empty, are independently normally distributed with mean 50 g50\,\mathrm{g} and standard deviation 2.0 g2.0\,\mathrm{g}.

(a) Find the variance of the masses of full cartons.

(2)

Small packets of rice are packed in boxes. The total masses of full boxes are normally distributed with mean 6730 g6730\,\mathrm{g} and standard deviation 15.0 g15.0\,\mathrm{g}. The masses of the boxes and cartons are distributed independently of each other.

(b) Find the probability that the mass of a randomly chosen full carton is more than three times the mass of a randomly chosen full box.

(5)
题目中文翻译

大袋大米装入纸箱，每个纸箱装入随机选取的 2020 袋大米。这些大袋大米的质量服从均值为 1010 g1010\,\mathrm{g}、标准差为 3.4 g3.4\,\mathrm{g} 的正态分布。空纸箱的质量独立地服从均值为 50 g50\,\mathrm{g}、标准差为 2.0 g2.0\,\mathrm{g} 的正态分布。

(a) 求装满大米的纸箱质量的方差。

(2)

小袋大米装入盒子。装满大米的盒子的总质量服从均值为 6730 g6730\,\mathrm{g}、标准差为 15.0 g15.0\,\mathrm{g} 的正态分布。盒子的质量与纸箱的质量相互独立。

(b) 随机选取一个装满大米的纸箱和一个装满大米的盒子，求纸箱的质量大于盒子质量三倍的概率。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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满箱质量为20个独立小包质量加空箱质量，方差按独立分量相加。

答题过程

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Let CC be the mass of a full carton. Its variance is

Var(C)= 20(3.4)2+(2.0)2= 231.2+4=235.2 g2.\begin{align*} \mathrm{Var}(C)=&\,20(3.4)^2+(2.0)^2\\[4mm] =&\,231.2+4=\boxed{235.2\text{ g}^2}. \end{align*}

(b)

解法一

思路

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先求满纸箱均值20250，再建立满纸箱减3倍满小盒的正态分布。3倍同一个盒子的质量使方差乘9。

答题过程

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The full carton has mean

E(C)=20(1010)+50=20250 g.\mathrm{E}(C)=20(1010)+50=20250\text{ g}.

Let BB be the full-box mass. By independence,

D=C−3B.D=C-3B.

Then

D∼ N(20250−3(6730), 235.2+32(15)2)= N(60,2260.2).\begin{align*} D\sim&\,\mathrm{N}\bigl(20250-3(6730),\ 235.2+3^2(15)^2\bigr)\\[4mm] =&\,\mathrm{N}(60,2260.2). \end{align*}

Therefore

P(C>3B)= 1−Φ(0−602260.2)= 0.896535…= 0.897(3 s.f.).\begin{align*} \mathrm{P}(C>3B)=&\,1-\Phi\bigg(\frac{0-60}{\sqrt{2260.2}}\bigg)\\[4mm] =&\,0.896535\ldots\\[4mm] =&\,\boxed{0.897}\quad\text{(3 s.f.).} \end{align*}